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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    The sum \[=\frac{3\times {{1}^{3}}}{{{1}^{2}}}+\frac{5\times \left( {{1}^{3}}+{{2}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}}+\frac{7\times \left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}} \right)}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+......\] [JEE Main 10-4-2019 Morning]

    A) 660

    B) 620

    C) 680                 

    D) 600

    Correct Answer: A

    Solution :

                \[{{T}_{n}}=\frac{(3+(n-1)\times 2)\left( {{1}^{3}}+{{2}^{3}}+...+{{n}^{3}} \right)}{\left( {{1}^{2}}+{{2}^{2}}+...+{{n}^{2}} \right)}\]           \[=\frac{3}{2}n(n+1)=\frac{n(n+1)(n+2)-(n-1)n(n+1)}{2}\]           \[\Rightarrow {{S}_{n}}=\frac{n(n+1)(n+2)}{2}\]\[\Rightarrow {{S}_{10}}=660\]       


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